Q1 Inner Join (3 points) Perform an inner join between the customers and orders datasets.

How many rows are in the result?

4 rows

Why are some customers or orders not included in the result?

An inner join only includes rows where there is a match in both datasets, which is why some customers (David & Eve) are not included.

Display the result
inner_join(customers, orders, by = "customer_id")
## # A tibble: 4 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300

Q2 Left Join (3 points) Perform a left join with customers as the left table and orders as the right table.

How many rows are in the result?

6 rows

Explain why this number differs from the inner join result.

This is different than inner join because left join displays all rows, even if they do not have a match in both datasets. So, David and Eve appear with an NA result in the orders column.

Display the result
left_join(customers, orders, by = "customer_id")
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA

Q3 Right Join (3 points) Perform a right join with customers as the left table and orders as the right table.

How many rows are in the result?

6 rows

Which customer_ids in the result have NULL for customer name and city? Explain why.

The 5th and 6th customers have NULL. This is because they exist in the orders table but do not exist in the customers table. Since there’s no matching customer, their details are missing.

Display the result
right_join(customers, orders, by = "customer_id")
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           6 <NA>    <NA>             105 Camera     600
## 6           7 <NA>    <NA>             106 Printer    150

Q4 Full Join (3 points) Perform a full join between customers and orders.

How many rows are in the result?

8 rows

Identify any rows where there’s information from only one table. Explain these results.

David and Eve do not have order information, and customers 7 and 8 do not have names or cities. This is because a full join includes all records from both tables. Some customers don’t have orders, and some orders don’t have matching customers, leading to NA values in the result.

Display the result
full_join(customers, orders, by = "customer_id")
## # A tibble: 8 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA
## 7           6 <NA>    <NA>             105 Camera     600
## 8           7 <NA>    <NA>             106 Printer    150

Q5 Semi Join (3 points) Perform a semi join with customers as the left table and orders as the right table.

How many rows are in the result?

3 rows

How does this result differ from the inner join result?

This result differs from the inner join result because the semi join only returns customers who have at least one order.The semi join has fewer rows than the inner join because it keeps only unique customers who placed orders.

Display the result
semi_join(customers, orders, by = "customer_id")
## # A tibble: 3 × 3
##   customer_id name    city       
##         <dbl> <chr>   <chr>      
## 1           1 Alice   New York   
## 2           2 Bob     Los Angeles
## 3           3 Charlie Chicago

Q6 Anti Join (3 points) Perform an anti join with customers as the left table and orders as the right table.

Which customers are in the result?

David and Eve

Explain what this result tells you about these customers.

These customers are in the result because they have not yet made a purchase.

Display the result
anti_join(customers, orders, by = "customer_id")
## # A tibble: 2 × 3
##   customer_id name  city   
##         <dbl> <chr> <chr>  
## 1           4 David Houston
## 2           5 Eve   Phoenix

Q7 Practical Application (4 points) Imagine you’re analyzing customer behavior.

Which join would you use to find all customers, including those who haven’t placed any orders? Why?

The only customers that count are the ones that appear in the customer table. To find these customers, I would use left join because this includes all customers, and those without orders will still appear, with their order details as NA.

Which join would you use to find only the customers who have placed orders? Why?

To find only customers who have placed orders, I would use inner join because this will return only customers who have matching orders, and exclude those who haven’t placed any.

Write the R code for both scenarios.
left_join(customers, orders, by = "customer_id")
## # A tibble: 6 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300
## 5           4 David   Houston           NA <NA>        NA
## 6           5 Eve     Phoenix           NA <NA>        NA
inner_join(customers, orders, by = "customer_id")
## # A tibble: 4 × 6
##   customer_id name    city        order_id product amount
##         <dbl> <chr>   <chr>          <dbl> <chr>    <dbl>
## 1           1 Alice   New York         101 Laptop    1200
## 2           2 Bob     Los Angeles      102 Phone      800
## 3           2 Bob     Los Angeles      104 Desktop   1500
## 4           3 Charlie Chicago          103 Tablet     300

Q8 Challenge Question (3 points) Create a summary that shows each customer’s name, city, total number of orders, and total amount spent. Include all customers, even those without orders. Hint: You’ll need to use a combination of joins and group_by/summarize operations.

The only customers that count are the ones that appear in the customer table.

customer_summary <- customers %>%
  left_join(orders, by = "customer_id") %>%
  group_by(customer_id, name, city) %>%
  summarize( 
    total_orders = n(),
    total_spent = sum(amount, na.rm = TRUE),
    .groups = "drop")
head(customer_summary)
## # A tibble: 5 × 5
##   customer_id name    city        total_orders total_spent
##         <dbl> <chr>   <chr>              <int>       <dbl>
## 1           1 Alice   New York               1        1200
## 2           2 Bob     Los Angeles            2        2300
## 3           3 Charlie Chicago                1         300
## 4           4 David   Houston                1           0
## 5           5 Eve     Phoenix                1           0